Question

If the local maximum value of the function f x)=\((\frac{√3e}{2 sin x} ) sin^2x ,\) \(x∈(0 ,\frac{π}{2}) \), is k/e , then  ( \(\frac{k}{ e}\) ) 8 + \(\frac{k^8}{e^ 5}\) + k ⁸  is equal to

• A. e⁵+e⁶+e¹¹
• B. e³+e⁶+e¹⁰
• C. e³+e⁵+e¹¹
• D. e³+e⁶+e¹¹

Hint:

To find the maximum or minimum value of a function, take the derivative and set it to zero. Remember the logarithmic differentiation technique for functions of the form f(x)g(x).

Answer 1

The Correct Option is D

Let:

\( y = \left(\frac{\sqrt{3}e}{2\sin x}\right)^{\sin^2 x} \)

Take the natural logarithm on both sides:

\( \ln y = \sin^2 x \cdot \ln\left(\frac{\sqrt{3}e}{2\sin x}\right) \)

Differentiate both sides with respect to \( x \):

\( \frac{1}{y} \cdot \frac{dy}{dx} = \ln\left(\frac{\sqrt{3}e}{2\sin x}\right) \cdot 2\sin x \cos x + \sin^2 x \cdot \frac{2\sin x \cdot \sqrt{3}e - \sqrt{3}e \cdot 2\cos x}{2 \cdot \sqrt{3}e} \)

Simplify the derivative:

\( \frac{dy}{dx} = y \cdot \left[\ln\left(\frac{\sqrt{3}e}{2\sin x}\right) \cdot 2\sin x \cos x - \sin x \cos x\right] \)

For local maxima or minima, set \( \frac{dy}{dx} = 0 \):

\( \ln\left(\frac{\sqrt{3}e}{2\sin x}\right) \cdot 2\sin x \cos x - \sin x \cos x = 0 \)

Factorize:

\( \sin x \cos x \cdot \left[2\ln\left(\frac{\sqrt{3}e}{2\sin x}\right) - 1\right] = 0 \)

For non-zero solutions:

\( \ln\left(\frac{\sqrt{3}e}{2\sin x}\right) = \frac{1}{2} \)

\( \frac{3e}{4\sin^2 x} = e^{1} \Rightarrow \frac{3e}{4\sin^2 x} = e \Rightarrow \sin^2 x = \frac{3}{4} \)

Hence:

\( \sin x = \frac{\sqrt{3}}{2} \quad \text{(as } x \in (0, \pi/2) \text{)} \)

The corresponding local maximum value of \( y \) is:

\( y = \left(\frac{\sqrt{3}e}{\sqrt{3}}\right)^{3/4} \)

Equating powers:

\( \left(\frac{k}{e}\right)^{8} = e^{3/8} = \frac{k}{e} \)

Using the given conditions:

\( k^8 = e^{11} \)

\( \left(\frac{k}{e}\right)^8 + \frac{k^8}{e^5} + k^8 = e^3 + e^6 + e^{11} \)